Question about the sum of chain groups What's the difference between $C_n(A + B)$ (that is, $C_n(A)+C_n(B)$ in $C_n(X)$) and $C_n(A) \oplus C_n(B)$ where $A$ and $B$ are subspaces of a topological space $X$? They're the same sets, right?
Thanks for your help.
 A: The point is that $C(A + B)$ is the subgroup of $C(A \cup B)$ generated by $C(A)$ and $C(B)$ while $C(A) \oplus C(B)$ is the abelian group consisting of pairs of elements $(c_1,c_2)$ with $c_1 \in C(A)$ and $c_2 \in C(B)$. The elements that $C(A)$ and $C(B)$ have in common (that is: exactly the elements of $C(A \cap B)$) occur twice in the direct sum $C(A) \oplus C(B)$: once as $(c,0)$ and once as $(0,c)$. On the other hand $C(A + B)$ is by definition the subgroup of all elements $c \in C(A \cup B)$ that can be written as the sum $c = c_1 + c_2$ with $c_1 \in C(A)$ and $c_2 \in C(B)$. In particular $C(A) \oplus C(B)$ and $C(A + B)$ are different (unless $A \cap B = \emptyset$).

Since I told you many times already that you should try and draw some pictures, here's how I would "see" Mayer-Vietoris:
Since the boundary map of $C(A \cup B)$ restricts to $C(A + B)$ (why?), we see that $C(A + B)$ is a chain complex.
Note that it is not all of $C(A \cup B)$ since a simplex $\sigma$ in $A \cup B$ can look as follows:

However, after (possibly a few) barycenric subdivisions,  each such simplex will be decomposed into (more formally: chain equivalent to a sum of) simplices that lie entirely inside either $A$ or $B$. And that's essentially why $H(C(A+B)) = H(C(A\cup B))$.

The precise proof is of course a bit more tedious, but that's essentially what's going on in Hatcher's Proposition 2.21.
Now we have a homomorphism $C(A) \oplus C(B) \to C(A + B)$ given by $(c_1, c_2) \mapsto c_1 + c_2$. This map is onto by definition of $C(A+B)$.
What is its kernel? 
You should convince yourself that the kernel consists precisely of the elements of the form $(c,-c)$ with $c \in C(A \cap B)$. In other words, we have a short exact sequence of chain complexes:
$$0 \;\xrightarrow{\phantom{blablabla}} \; C(A \cap B) \; \xrightarrow[\phantom{blablabla}]{c \mapsto (c,-c)}\; C(A) \oplus C(B) \;
\xrightarrow[\phantom{blablabla}]{(c_1,c_2) \mapsto c_1 + c_2}\; C(A+B)\;
\xrightarrow[\phantom{blablabla}]{}\; 0$$
But this short exact sequence exhibits $C(A + B)$ as (isomorphic to) the quotient $\left(C(A) \oplus C(B)\right)\; / \; C(A\cap B)$. The Mayer-Vietoris sequence is then the long exact sequence in homology associated to this short exact sequence of complexes.
So the "difference" between $C(A + B)$ and $C(A) \oplus C(B)$ is precisely measured by $C(A \cap B)$: Chains lying in $A \cap B$ can belong to either $C(A)$ or $C(B)$. So no, $C(A+B)$ and $C(A) \oplus C(B)$ not the same sets.
A simple situation that shows intuitively what is going on here is: Let $X = \mathbb{R}^3$ and consider the subspaces $U = \langle e_1,e_2 \rangle$ and $V= \langle e_2, e_3 \rangle$ each spanned by two out of three vectors of the standard basis. Then $X = U + V$ while the (exterior) direct sum $U \oplus V$ has four dimensions. The superfluous dimension comes from the fact that $U \cap V = \langle e_2 \rangle$ is non-zero and we have a similar short exact sequence $0 \to U \cap V \to U \oplus V \to U + V \to 0$ as before.
